integral from 0 to pi of 3sin^2(xco)s^4x

解

\int_{0}^{π} 3 sin^{2} (x co) s^{4} x d x

\frac{3 s ^{4} ( 2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co ) + 1 )}{8 c ^{2} o ^{2}}

解答ステップ

\int_{0}^{π} 3 sin^{2} (x co) s^{4} x d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 s^{4} \cdot \int_{0}^{π} sin^{2} (co x) x d x

u置換積分法を適用する

= 3 s^{4} \cdot \int_{0}^{co π} \frac{u sin ^{2} ( u )}{c ^{2} o ^{2}} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 s^{4} \frac{1}{c ^{2} o ^{2}} \cdot \int_{0}^{co π} u sin^{2} (u) d u

部分積分を適用する

= 3 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u]_{0}^{π co}

\int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u = \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))

= 3 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))]_{0}^{π co}

簡素化

3 s^{4} \frac{1}{c ^{2} o ^{2}} [\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))]_{0}^{π co} : \frac{3 s ^{4}}{c ^{2} o ^{2}} [\frac{2 u ^{2} - 2 u sin ( 2 u ) - cos ( 2 u )}{8}]_{0}^{π co}

= \frac{3 s ^{4}}{c ^{2} o ^{2}} [\frac{2 u ^{2} - 2 u sin ( 2 u ) - cos ( 2 u )}{8}]_{0}^{π co}

限度を計算する:

\frac{2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co )}{8} + \frac{1}{8}

= \frac{3 s ^{4}}{c ^{2} o ^{2}} (\frac{2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co )}{8} + \frac{1}{8})

簡素化

= \frac{3 s ^{4} ( 2 π ^{2} c ^{2} o ^{2} - 2 π co sin ( 2 π co ) - cos ( 2 π co ) + 1 )}{8 c ^{2} o ^{2}}